1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Change of variables for multiple integrals (3)

  1. Jan 24, 2008 #1
    Q1: Let S be the region in the first quadrant bounded by the curves xy=1, xy=3, x2 - y2 = 1, and x2 - y2 = 4. Compute
    ∫∫(x2 + y2)dA.
    (Hint: Let G(x,y)=(xy, x2 - y2). What is |det DG|?)


    I don't understand the third and fourth equalities in the solutions. Can someone kindly explain?
  2. jcsd
  3. Jan 24, 2008 #2


    User Avatar
    Science Advisor

    If you don't understand why
    [tex]\int\int_S(x^2+ y^2)dA= \frac{1}{2}\int\int_S |det DG| dA[/tex]
    then you have missed the whole point of this section! |det DG| is the "Jacobian" and |det DG| dA is the differential in terms of the new variables u= xy, v= x2- y2.
    [tex]det DG= \left[\begin{array}{cc}\frac{\partial xy}{\partial x} & \frac{\partial x^2- y^2}{\partial x} \\ \frac{\partial xy}{\partial y} & \frac{\partial x^2- y^2}{\partial y}\end{array}\right]= \left[\begin{array}{cc}y & 2x \\ x & -2y\end{array}\right]= -2(x^2+ y^2)[/itex]
    as they say so |det dG|= 2(x2+ y2). You already have x2+ y2 in the integral so you only need to multiply inside the integral by 2 and multiply outside the integral by 1/2 to get [itex]1/2 \int\int |det dG|dA[/itex].

    Now, for the limits of integration. Your region is bounded by xy= 1, xy= 3, x2- y2= 1 and x2- y2= 4. That, of course, was the reason for the choice of G. With u= xy, u= xy= 1 and u= xy= 3 are the u limits of integration. With v= x2- y2, v= x2- y2= 1 and v= x2- y2= 4 are the v limits of integration.
  4. Jan 25, 2008 #3
    Change of variables theorem says
    =∫∫f(G(u))|det DG| dudv ?

    So you should add the Jaconbian factor, but why are they moving it away? (it seems that they are doing it the oppposite way than the theorem states)

    For the fourth equality, I don't understand why |det DG| disappeared.
  5. Jan 25, 2008 #4


    User Avatar
    Science Advisor

    You are going in the wrong direction.
    Suppose you had the problem [itex]\int x dx[/itex] (yes, I know that's trivial!) and you decide to make the substitution u= x2 (yes, I know that's a peculiar choice!). Then du= 2x dx so (1/2)du= xdx. Since you already have "xdx" in the integral, in terms of u, you have just [itex](1/2)\int du= (1/2)u+ C= (1/2)x^2+ C[/itex]. The "x" inside the integral was "absorbed" into the du.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Change of variables for multiple integrals (3)